Refractory MgO and Non-stoichiometric TiO
A ceramic plant supplies magnesium oxide powder for refractory bricks. X-ray diffraction shows a face-centred cubic NaCl-type lattice with edge length a = 4.21 Å and four formula units per unit cell (M(MgO) = 40.30 g mol⁻¹). A helium pycnometer gives a density of 3.45 g cm⁻³, whereas the buyer's specification quotes 3.58 g cm⁻³. A second supplier sends 'TiO' with the same NaCl-type structure and a = 4.18 Å; its measured density is 4.0% higher than the value calculated from the unit cell, and the powder is dark grey instead of white. The plant chemist must decide whether each batch is stoichiometric, metal-deficient or metal-excess, because non-stoichiometry changes sintering behaviour and electrical conductivity. Ionic radii available: Mg²⁺ 0.72 Å, Ti²⁺ 0.86 Å, O²⁻ 1.40 Å. The customer will reject any batch whose density deviates by more than 2% from the ideal value.
1. Calculate the theoretical density of MgO from the unit-cell data, compare it with the measured value, and decide which defect model is consistent. Estimate the fraction of vacant sites and explain why Frenkel defects would not explain the observation.
Read the worked answer
Use (d = \frac{ZM}{N_A a^3}). Here (a = 4.21\times10^{-8}) cm, so (a^3 = 7.462\times10^{-23}) cm³ and (N_Aa^3 = 44.93) cm³ mol⁻¹. Thus
[d = \frac{4\times40.30}{44.93} = 3.59\ \text{g cm}^{-3}.]
The measured value is 3.45 g cm⁻³, i.e. (3.45/3.59 = 0.961), about 3.9% low. A lower density at essentially unchanged unit-cell volume means fewer formula units per cell, not a larger cell. That is the signature of Schottky defects: paired cation and anion vacancies. The fraction is (x \approx 0.039), so roughly 3.9% of the Mg²⁺ and 3.9% of the O²⁻ sites are vacant. Frenkel defects move an ion from a lattice site to an interstitial site: the same mass remains in the same cell volume, so the density stays at the ideal value. They cannot produce a 3.9% drop. A lower-molar-mass impurity could also lower density, but the XRD pattern is single-phase MgO, so the Schottky interpretation is self-consistent. This batch also fails the customer's 2% limit.
2. Using the radius ratio, predict the coordination of Mg²⁺ and check whether it agrees with the NaCl-type structure. Would a radius ratio of 0.35 change your conclusion?
Read the worked answer
[\frac{r_+}{r_-} = \frac{0.72}{1.40} = 0.514.]
A ratio between 0.414 and 0.732 predicts an octahedral hole and coordination number 6. NaCl-type structures have 6:6 coordination, so the prediction agrees with the diffraction result: Mg²⁺ occupies octahedral holes formed by O²⁻. If the ratio were 0.35, it would fall in the 0.225–0.414 range, favouring a tetrahedral hole with 4:4 coordination and a ZnS-type structure. That would contradict the observed NaCl-type pattern. The radius-ratio rule is a guide rather than a strict law: polarisation, covalency and temperature can shift the boundary, so the XRD result remains the decisive evidence.
3. The TiO batch has a density 4.0% above the calculated value. Decide whether it is metal-excess or metal-deficient, identify a plausible defect, and explain the dark colour and conductivity.
Read the worked answer
For ideal TiO with (Z=4), (M = 63.87) g mol⁻¹ and (a = 4.18\times10^{-8}) cm:
[d = \frac{4\times63.87}{6.022\times10^{23}\times(4.18\times10^{-8})^3} = \frac{255.5}{43.98} = 5.81\ \text{g cm}^{-3}.]
The measured value is about 6.04 g cm⁻³. At essentially fixed cell volume, a higher mass per cell means extra titanium relative to oxygen: the sample is metal-excess, not metal-deficient. A plausible defect is interstitial Ti (a cation Frenkel-type defect), possibly together with oxygen vacancies. Charge balance converts some Ti²⁺ to Ti³⁺ and releases electrons. Metal deficiency would remove cations and lower the density, so it is excluded. The released electrons occupy anion vacancies (F-centres) or Ti 3d states, giving the dark grey colour and n-type electronic conductivity. Both the density and the colour therefore point the same way.